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Photoelectric Effect

by Noir
Tags: effect, photoelectric
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Noir
#1
Nov8-09, 10:13 PM
P: 27
1. The problem statement, all variables and given/known data
When 520nm light falls on a metal, the current through the circuit is brought to zero by a reverse voltage of 1.13V. What is the threshold frequency for this metal.


2. Relevant equations
E = h.f


3. The attempt at a solution
Using the above formula I get an threshold frequency of 2.72 x 10^14 Hz. Except here's my question, the reverse voltage doesn't have anything to do with the work function if it brings the current down to zero. Or does it, can someone please explain the process? I'm just confused by the term reverse voltage, or stopping voltage.

Thanks for any help :)
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rl.bhat
#2
Nov8-09, 10:29 PM
HW Helper
P: 4,433
The maximum kinetic energy of the ejected electron is given by the stopping potential.
According to photoelectric effect
KE = hν - φ where φ is the work function.
Noir
#3
Nov8-09, 10:59 PM
P: 27
Thanks for your reply, but I'm still unclear on how the reverse voltage, or stopping voltage relates to the equarion.

rl.bhat
#4
Nov8-09, 11:15 PM
HW Helper
P: 4,433
Photoelectric Effect

Kinetic energy = e*V, where e is the charge on electron and V is the stopping voltage.


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